AB is a diameter of the circle with center O. The tangent at a point P meets AB produced in Q, ∠PAQ = 34°, find angles ∠PBA and ∠PQA.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠APB = 90°

In △ APB,

⇒ ∠APB + ∠PAB + ∠PBA = 180°

⇒ 90° + 34° + ∠PBA = 180°

⇒ ∠PBA = 180° - 90° - 34° = 56°.

We know that,

Angle subtended by an arc at the center of the circle is twice the angle subtended by it on the circumference of the circle.

∴ ∠POB = 2∠PAB = 2 × 34° = 68°.

In △ POQ,

⇒ ∠OPQ + ∠PQO + ∠POQ = 180°

⇒ 90° + 68° + ∠PQO = 180°

⇒ 158° + ∠PQO = 180°

⇒ ∠PQO = 180° - 158° = 22°.

From figure,

⇒ ∠PQA = ∠PQO = 22°.

Hence, ∠PBA = 56° and ∠PQA = 22°.